Does this property characterize abelian groups?
$begingroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
Let $G$ be a group. Suppose there exists an integer $k>1$ and a non-identity permutation $pi in S_k$ such that for all $x_1, x_2 cdots x_k neq mathbf{1} in G$ we have that $x_1x_2x_3 cdots x_k = x_{pi(1)}x_{pi(2)}x_{pi(3)} cdots x_{pi(k)}$. Must $G$ be abelian?
This question is motivated by this. I've attempted to use Andrés technique in the linked post again here, but to no avail. I've also attempted to look at classical non-abelian groups like $S_3$ and $GL_n(mathbf{C})$ for counterexamples but also to no avail.
If this is false, is it possible that there's some conditions on the permutation $pi$ which makes this true? Because, again looking over the linked post, it is clear that some permutations force abelianness.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
edited Jan 2 at 23:29
MathematicsStudent1122
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
asked Jan 2 at 22:44
MathematicsStudent1122MathematicsStudent1122
8,67622467
8,67622467
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,808929
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060064%2fdoes-this-property-characterize-abelian-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,808929
$endgroup$
add a comment |
$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,808929
$endgroup$
add a comment |
$begingroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,808929
$endgroup$
Yes. By induction on $kge 2$; $k=2$ is clear.
Suppose $kge 3$. If $pi(k)=k$ then it's clear by induction. Otherwise, write $j=pi(k)$. Write the equality with $x_j=x_k$. It yields $x_1dots x_{k-1}=x_{pi(1)}dots x_{pi(k-1)}$; write $rho(i)=pi(i)$ for $1le ile k-1$, $ineqpi^{-1}(k)$, $rho(pi^{-1}(k))=j$ ($rho$ is the 1st return map on ${1,dots,k-1}$); then $x_1dots x_{k-1}=x_{rho(1)}dots x_{rho(k-1)}$ for all $x_1,dots,x_{k-1}$ not 1. Hence, if $rho$ is not the identity, we are done.
Then we see that $rho$ is the identity if and only if $pi$ is the transposition $tau_{j,k}$. Since the case when $pi(1)=1$ is also clear, it remains to assume that $pi=tau_{1,k}$. So we have $x_1dots x_k=x_kx_2dots x_{k-1}x_1$ for all $x_ineq 1$. If we find $x_2,dots,x_{k-1}$ not 1 whose product is 1, we are done. This is obviously doable if $k$ is even (with $x_{2i+1}=x_{2i}^{-1}$ for $1<2i<k$, the case $|G|=1$ being trivially excluded); this is doable for $k=5$ as soon as $|G|ge 3$ and then for $kge 5$ odd ($|G|=2$ being trivial too). Finally, the remaining case is $k=3$ and the equality $x_1x_2x_3=x_3x_2x_1$, and this case was done in the linked post.
answered Jan 2 at 23:32
YCorYCor
7,808929
answered Jan 2 at 23:32
YCorYCor
7,808929
answered Jan 2 at 23:32
YCorYCor
7,808929
answered Jan 2 at 23:32
YCorYCor
7,808929
7,808929
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060064%2fdoes-this-property-characterize-abelian-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
